AOT - Cell Potentials, Gibbs Free Energy, and the Equilibrium Constant

Cell Potentials, Gibbs Free Energy, and the Equilibrium Constant

The previous lesson introduced the idea of electrochemical cell potentials, how to calculate them, and their ability to determine spontaneity. The latter is able to be determined by evaluating the sign of the E_cell value. If the cell potential is positive, then the reaction proceeds spontaneously in the direction that it is written. As has also been noted in a prior module, the Gibbs free energy value can also determine spontaneity with a negative value indicating spontaneity in that case. Since each of these properties has the ability to determine this spontaneity, it stands to reason that there must be some type of relationship between the two. Indeed, there is a relationship as shown below:

$LaTeX: \Delta G=-nFE$ $\Delta G=-nFE$

$LaTeX: \Delta G^{^{\circ}}=-nFE^{^{\circ}}$ $\Delta G^{^{\circ}}=-nFE^{^{\circ}}$ (std. conditions)

The value for n in each of these relationships refers to the number of electrons transferred during the redox process and F refers to Faraday's constant whose value is:

$LaTeX: F=96,485\frac{}{}\frac{J}{V\cdot mol}$ $F=96,485\frac{}{}\frac{J}{V\cdot mol}$

Given the redox reaction below, this relationship allows us to directly calculate the Gibbs free energy change by using standard reduction potentials. The pertinent reduction potentials are also provided, but an extensive list of reduction potentials Links to an external site. can be accessed from the OpenStax textbook.

2 Fe⁺³ (aq) + 2 I^- (aq) $LaTeX: \longrightarrow$ $\longrightarrow$ 2 Fe⁺² (aq) + I₂ (s)

Reduction half-reaction (cathode): 2 Fe⁺³ (aq) + 2 e^- $LaTeX: \longrightarrow$ $\longrightarrow$ 2 Fe⁺² (aq) $LaTeX: E^{^{\circ}}_{red}$ $E^{^{\circ}}_{red}$ = +0.771 V
Oxidation half-reaction (anode): 2 I^-(aq) $LaTeX: \longrightarrow$ $\longrightarrow$ I₂ (s) + 2 e^- $LaTeX: E^{^{\circ}}_{red}$ $E^{^{\circ}}_{red}$ = +0.536 V
The number of electrons transferred is two (n = 2)

The first step is to calculate the cell potential (E_cell) for this reaction:

$LaTeX: E_{cell}^{^{\circ}}=E_{red}^{^{\circ}}\left(cathode\right)-E_{red}^{^{\circ}}\left(anode\right)$ $E_{cell}^{^{\circ}}=E_{red}^{^{\circ}}\left(cathode\right)-E_{red}^{^{\circ}}\left(anode\right)$

$LaTeX: E_{cell}^{^{\circ}}=$ $E_{cell}^{^{\circ}}=$ 0.771 V LaTeX: - $-$ 0.536 V

$LaTeX: E_{cell}^{^{\circ}}=$ $E_{cell}^{^{\circ}}=$ 0.235 V

A positive cell potential is calculated which means that a negative Gibbs free energy change should be expected to maintain a consistent conclusion that this reaction is spontaneous in the direction that it is written.

$LaTeX: \Delta G^{^{\circ}}=-nFE^{^{\circ}}$ $\Delta G^{^{\circ}}=-nFE^{^{\circ}}$

$LaTeX: \Delta G^{^{\circ}}=-\left(2\right)\left(96485\right)\left(0.235\right)$ $\Delta G^{^{\circ}}=-\left(2\right)\left(96485\right)\left(0.235\right)$

$LaTeX: \Delta G^{^{\circ}}=$ $\Delta G^{^{\circ}}=$ -45300 J/mol OR -45.3 kJ/mol

You Try It!

In the following self-assessment activity, determine the Gibbs free energy change for the reaction. Click on the plus sign to check your answer!

Cell Potential and the Equilibrium Constant

Given the previously established relationship between the Gibbs free energy change and the equilibrium constant as well as the relationship between Gibbs free energy and cell potential described above, the following connection between cell potential and equilibrium constant can be made:

$LaTeX: \Delta G^{^{\circ}}=-RTln\left(K\right)$ $\Delta G^{^{\circ}}=-RTln\left(K\right)$ $LaTeX: \Delta G^{^{\circ}}=-nFE^{^{\circ}}$ $\Delta G^{^{\circ}}=-nFE^{^{\circ}}$

$LaTeX: -RTln\left(K\right)$ $-RTln\left(K\right)$ $LaTeX: =-nFE^{^{\circ}}$ $=-nFE^{^{\circ}}$

$LaTeX: E^{^{\circ}}=\frac{RT}{nF}ln\left(K\right)$ $E^{^{\circ}}=\frac{RT}{nF}ln\left(K\right)$

The connections between these three properties can be visualized using the diagram below.

Using this relationship, the equilibrium constant for the reaction between Sn (IV) and U (III) ions at 600 K can be determined. Once again the table of standard reduction Links to an external site. potentials will be referenced in order to determine the standard cell potential.

3 Sn⁺⁴ (aq) + 2 U (s) $LaTeX: \longrightarrow$ $\longrightarrow$ 3 Sn⁺² (aq) + 2 U⁺³ (aq)

First, begin by splitting the overall reaction into its two constituent half-reactions to calculate the standard cell potential.

Reduction half-reaction: 3 Sn⁺⁴ (aq) + 6 e^- $LaTeX: \longrightarrow$ $\longrightarrow$ 3 Sn⁺² (aq) $LaTeX: E^{^{\circ}}_{red}$ $E^{^{\circ}}_{red}$ = 0.15 V
Oxidation half-reaction: 2 U (s) $LaTeX: \longrightarrow$ $\longrightarrow$ 2 U⁺³ (aq) + 6 e^- $LaTeX: E^{^{\circ}}_{red}$ $E^{^{\circ}}_{red}$ = -1.79 V
The number of electrons transferred is 6 (n = 6)

$LaTeX: E_{cell}^{^{\circ}}=$ $E_{cell}^{^{\circ}}=$ 0.15 V LaTeX: - $-$ (-1.79) V

$LaTeX: E_{cell}^{^{\circ}}=$ $E_{cell}^{^{\circ}}=$ 1.94 V

With this value in hand, the equilibrium constant can be calculated.

$LaTeX: E^{^{\circ}}=\frac{RT}{nF}ln\left(K\right)$ $E^{^{\circ}}=\frac{RT}{nF}ln\left(K\right)$

$LaTeX: 1.94=\frac{\left(8.314\right)\left(600\right)}{\left(6\right)\left(96485\right)}ln\left(K\right)$ $1.94=\frac{\left(8.314\right)\left(600\right)}{\left(6\right)\left(96485\right)}ln\left(K\right)$

225.14 = LaTeX: ln(K) $ln(K)$

$LaTeX: K=5.98\times10^{97}$ $K=5.98\times10^{97}$

The positive standard cell potential indicates that the reaction is spontaneous, and the calculation of K reveals something further. With an equilibrium constant on the order of 10⁹⁷, it is quite clear that this reaction lies very far to the right indicating a highly spontaneous process.

You Try It!

In the following self-assessment activity, determine the equilibrium constant for the reaction. Click on the plus sign to check your answer!

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